The activity of a radioactive sample drops from $4000\,\text{decays/s}$ to $500\,\text{decays/s}$ in $30$ minutes. The half-life of the sample is

1. Activity $A(t)$ obeys the same exponential decay law as $N(t)$: $A(t) = A_0\,(1/2)^{t/T_{1/2}}$. 2. Find the ratio: $A(t)/A_0 = 500/4000 = 1/8$. 3. Set $(1/2)^{t/T_{1/2}} = 1/8 = (1/2)^3$. So $t/T_{1/2} = 3$. 4. With $t = 30$ minutes: $T_{1/2} = t/3 = 30/3 = 10$ minutes. 5. Step-by-step check: at $t = 0$, $A = 4000$; after $10$ min, $A = 2000$; after $20$ min, $A = 1000$; after $30$ min, $A = 500$. ✓ 6. Option A makes $t/T_{1/2} = 6$ giving a factor $1/64$, not $1/8$. Option C and D give wrong number of half-lives. _Source: NCERT Class 12 Physics Part 2, Ch 13, §13.6 (Radioactive Decay Law applied to activity), p. 8._